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A plant is running on 400 V, 50 Hz LT supply for 10 hours. The load comprises lighting load of 2.5kW @ 0.9 pf lagging on R-phase, heater load 5 kW @ UPF on Y-Phase and a 3-Phase SCIM drawing 5 kW @ 0.8 pf lagging. What is the total energy consumption by the loads?

A plant is running on 400 V, 50 Hz LT supply for 10 hours. The load comprises lighting load of 2.5kW @ 0.9 pf lagging on R-phase, heater load 5 kW @ UPF on Y-Phase and a 3-Phase SCIM drawing 5 kW @ 0.8 pf lagging. What is the total energy consumption by the loads? Correct Answer 125 units

Energy consumption:

Energy consumption is the use of power or energy of a system by making use of supply.

Energy = (Power x time) / 1000.

In electrical systems, Energy is calculated in kWh (kilo Watt-hours).

So, 1 unit of energy = 1kWh.

Power is in Watts 

Time in hours.

Calculation:

All loads connected for 10 hrs

load comprises lighting load of 2.5kW @ 0.9 pf lagging on R-phase

Energy consumed on R phase = 2.5kW × 10hr = 25kWh = 25 units

heater load 5 kW @ UPF on Y-Phase

Energy consumed on Y phase = 5kW × 10hr = 50kWh = 50 units

3-Phase SCIM drawing 5 kW @ 0.8 pf lagging on B phase

Energy consumed on B phase = 5kW × 10hr = 50kWh = 50 units

Total energy consumption by the loads = 25 + 50 + 50 = 125 units

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